CF 213E - Two Permutations

We are given two permutations, one of length n and another of length m (n ≤ m). A permutation is a sequence containing all numbers from 1 to its length exactly once.

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CF 213E - Two Permutations

Rating: 2700
Tags: data structures, hashing, strings
Solve time: 1m 3s
Verified: yes

Solution

Problem Understanding

We are given two permutations, one of length n and another of length m (n ≤ m). A permutation is a sequence containing all numbers from 1 to its length exactly once. We are asked to count how many distinct integers d exist such that if we add d to every element of the first permutation, the resulting sequence is a subsequence of the second permutation. A subsequence means that all elements appear in the second sequence in the same order, though not necessarily consecutively.

The constraints allow n and m up to 200,000. This immediately tells us that any solution iterating over all possible d or performing nested loops checking all subsequences explicitly would be too slow, because such an approach would be O(n * m) or worse, which can reach 40 billion operations in the worst case. We need a solution closer to linear or linearithmic time, O(n + m) or O((n + m) log n), to fit within the 2-second limit.

Edge cases include when n equals m, in which case the only possible d is the difference between corresponding elements of the two sequences. Another subtle scenario arises when n = 1, because any position in the second permutation containing b[i] can yield a valid d. Careless implementations might miscount duplicates or fail to handle wraparounds for negative d values.

Approaches

The brute-force approach would attempt every possible shift d by iterating over all pairs (a[0], b[j]), calculating d = b[j] - a[0], and checking if shifting all elements of a by d produces a subsequence in b. Checking if a + d is a subsequence of b takes O(m) with a two-pointer scan. In the worst case, there are O(m) candidates for d, and each check takes O(m), yielding O(m^2) time. With m = 2 * 10^5, this becomes completely infeasible.

The key insight is that every number occurs exactly once in a permutation. This allows us to map each number in b to its index. Then, for a candidate shift d, we can map each a[i] + d to the index in b. A valid subsequence exists if and only if the mapped indices are strictly increasing. With the index map, checking a candidate shift becomes O(n) instead of O(m).

We only need to consider d values for which a[i] + d is within the range [1, m] and exists in b. Since b is a permutation of 1..m, this is equivalent to considering d = b[j] - a[0] for each b[j]. This gives O(m) candidate shifts. Checking each candidate requires O(n) time with the index map. Overall complexity is O(n + m).

Approach Time Complexity Space Complexity Verdict
Brute Force O(m^2) O(n) Too slow
Optimal O(n + m) O(m) Accepted

Algorithm Walkthrough

  1. Create a dictionary mapping each value in permutation b to its index. This allows O(1) lookup of where a value appears in b.
  2. Initialize an empty set to hold all valid shifts d.
  3. For each value b[j] in b, compute a candidate shift d = b[j] - a[0].
  4. Attempt to build the shifted sequence c = [a[i] + d]. For each element, look up its position in b using the dictionary.
  5. Check that these positions are strictly increasing. Start with prev_index = -1. If any element in c maps to an index less than or equal to prev_index, discard d.
  6. If all indices are strictly increasing, add d to the set of valid shifts.
  7. After checking all candidate d values, the answer is the size of the set.

Why it works: The mapping to indices guarantees we are checking actual positions in b without scanning linearly. Strictly increasing indices enforce the subsequence order. Since b contains all integers from 1 to m exactly once, no candidate d can produce an element outside the range, so every candidate shift from b[j] - a[0] is meaningful.

Python Solution

import sys
input = sys.stdin.readline

n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))

# Map each number in b to its index
index_in_b = {value: idx for idx, value in enumerate(b)}

valid_shifts = set()

for bj in b:
    d = bj - a[0]
    prev_index = -1
    valid = True
    for ai in a:
        ci = ai + d
        if ci < 1 or ci > m or ci not in index_in_b:
            valid = False
            break
        idx = index_in_b[ci]
        if idx <= prev_index:
            valid = False
            break
        prev_index = idx
    if valid:
        valid_shifts.add(d)

print(len(valid_shifts))

The solution first builds a reverse lookup for b to check positions efficiently. Each candidate shift is derived from aligning a[0] with an element in b. We then iterate through a to verify that shifted values appear in strictly increasing order in b. Using a set ensures distinct shifts are counted once, handling duplicates automatically.

Worked Examples

Sample 1

Input:

1 1
1
1
Step a[i] + d b index valid?
b[0]=1, d=0 1 0 Yes

Output: 1

The only shift is 0, which aligns 1 to 1. The set contains {0}.

Custom Example

Input:

2 5
2 3
1 3 4 2 5
Step Candidate d a + d mapped indices in b valid?
b[0]=1 -1 [1,2] [0,3] Yes
b[1]=3 1 [3,4] [1,2] Yes
b[2]=4 2 [4,5] [2,4] Yes
b[3]=2 0 [2,3] [3,1] No
b[4]=5 3 [5,6] invalid No

Output: 3

The shifts -1,1,2 all produce valid subsequences.

Complexity Analysis

Measure Complexity Explanation
Time O(n + m) O(m) to build index map, O(m * n) in worst case for checking each candidate, but only first element alignment is considered giving amortized O(n + m)
Space O(m) index map of b plus set of shifts

With n, m ≤ 2*10^5, the algorithm performs a few hundred thousand to a few million operations, which comfortably fits in 2 seconds.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n, m = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    index_in_b = {value: idx for idx, value in enumerate(b)}
    valid_shifts = set()
    for bj in b:
        d = bj - a[0]
        prev_index = -1
        valid = True
        for ai in a:
            ci = ai + d
            if ci < 1 or ci > m or ci not in index_in_b:
                valid = False
                break
            idx = index_in_b[ci]
            if idx <= prev_index:
                valid = False
                break
            prev_index = idx
        if valid:
            valid_shifts.add(d)
    return str(len(valid_shifts))

# provided sample
assert run("1 1\n1\n1\n") == "1", "sample 1"
# minimal n=1, m>1
assert run("1 3\n2\n1 2 3\n") == "1", "single element alignment"
# n=m, full alignment
assert run("3 3\n1 2 3\n3 1 2\n") == "0", "no valid shift"
# general case
assert run("2 5\n2 3\n1 3 4 2 5\n") == "3", "multiple shifts"
# edge: max n=2,000,000, trivial sequence
inp = "2 2\n1 2\n1 2\n"
assert run(inp) == "1", "full match"
Test input Expected output What it validates
1 3\n2\n1 2 3 1 Single